Fe + H2SO4 --> FeSO4 + H2
Ta có nH2SO4= 11,2/56=0,2=nFe= nFeSO4
=> mFeSO4= 0,2.152=30,4 g
b, mdd H2SO4= 0,2.98.100/19,6=100 g
c, C%dd sau PỨ= 30,4.100/(11,2 + 100- 0,2.2)=27,4%
nFe = 11.2/56= 0.2 mol
Fe + H2SO4 --> FeSO4 + H2
0.2___0.2_______0.2____0.2
mH2SO4 = 0.2*98=19.6g
mddH2SO4 = 19.6*100/19.6=100g
mH2 = 0.4 g
mFeSO4 = 0.2*152 = 30.4 g
mdd sau phản ứng = 11.2 + 100 - 0.4 = 110.8g
C%FeSO4 = 30.4/110.8*100% = 27.44%
Fe + H2SO4 → FeSO4 + H2↑
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
a) Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,2\times152=30,4\left(g\right)\)
Theo pT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
b) Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,2\times98=19,6\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{19,6}{19,6\%}=100\left(g\right)\)
c) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)
Ta có: \(m_{dd}saupư=11,2+100-0,4=110,8\left(g\right)\)
\(C\%_{FeSO_4}=\frac{30,4}{110,8}\times100\%=27,44\%\)
a) PTHH: Fe + H2SO4 loãng \(\rightarrow\) FeSO4 + H2\(\uparrow\)
...............0,2......0,2.....................0,2..........0,2
nFe = \(\frac{11,2}{56}=0,2\left(mol\right)\)
m\(FeSO_4\) = 0,2.152 = 30,4 (g)
V\(H_2\) = 0,2.22,4 = 4,48 (l)
b) m\(H_2SO_4\) = 0,2.98 = 19,6(g)
=> mdd \(H_2SO_4\) = \(\frac{19,6}{19,6}.100\) = 100 (g)
c) m\(H_2\) = 0,2.2 = 0,4 (g)
mdd sau pứ = mFe + mdd \(H_2SO_4\) - m\(H_2\) = 11,2 + 100 - 0,4 = 110,8 (g)
=> C%\(FeSO_4\) = \(\frac{30,4}{110,8}.100\%\) = 27,44 %
\(Fe+H_2SO_4-->FeSO_4+H_2\)
0,2____0,2__________0,2_____0,2
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
a) \(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b) \(m_{d^2H_2SO_4}=\frac{0,2.98.100}{19,6}=100\left(g\right)\)
c) \(m_{d^2sau}=11,2+100-0,2.2=110,8\left(g\right)\)
=> \(C\%_{d^2sau}=\frac{30,4}{110,8}.100=27,44\%\)
